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Jul 7, 2010 at 17:32 answer added user7361 timeline score: 10
Jul 2, 2010 at 1:57 comment added Phil Isett Thanks for these replies. Maybe I should clarify: I'm not exactly looking for a proof of a fact for more general topological groups. (I'm really grateful for the example, though.) I just thought the key difference between the p-adics and Euclidean space in this regard has to do with the fact that convolutions (of, say, non-negative, compactly supported functions) really do expand support on R^n, but don't need to on the p-adics. I'd like to see a proof that highlights that feature of Euclidean space if there is one.
Jul 1, 2010 at 13:49 comment added BS. Yemon, I don't think this to be true : the inverse limit of $2\times : \mathbb T \to \mathbb T$ (a so called solenoid) seems to be connected. See en.wikipedia.org/wiki/Solenoid_(mathematics).
Jun 30, 2010 at 23:38 history edited Phil Isett CC BY-SA 2.5
Clarified question
Jun 30, 2010 at 6:54 comment added Yemon Choi By the way, there ought to be a classification of connected locally compact abelian groups in Hewitt & Ross volume 1 (though that may not be the most effective or convenient source). IIRC, they are all of the form ${\mathbb R}^n \times {\mathbb T}^k$.
Jun 30, 2010 at 6:52 comment added Yemon Choi Scott, I don't think that's what's being asked? As I understand it, the question is something like the following: is there a "connectedness-driven proof" that, when G is abelian connected and locally compact, and f is a function vanishing on an open subset of G, then the FT of f cannot have compact support.
Jun 30, 2010 at 2:55 comment added S. Carnahan I don't think connectedness alone kills compact support in frequency space, since you can have compactly supported functions on $\mathbb{Z}$ or $S^1$ (or products of these) with compactly supported Fourier transforms.
Jun 30, 2010 at 0:43 history asked Phil Isett CC BY-SA 2.5